Q9Q - Fundamentals Of Physics

Question

An electron that is trapped in a one-dimensional infinite potential well of width $\mathbf{L}$is excited from the ground state to the first excited state. Does the excitation increase, decrease, or have no effect on the probability of detecting the electron in a small length of the x axis (a) at the center of the well and (b) near one of the well walls?

Step-by-Step Solution

Answer

(a)The probability is decreases as the electron go from the ground state to first excited state at the centre.

(b)The probability is increases as the electron go from the ground state to first excited near the one of the well walls.

Step 1: Write the given data from the question.

The width of the one-dimensional infinite potential well is $L$.

Step 2: Determine the formulas to calculate the effect of excitation on the probability.

The expression to calculate the wave function for the electron is given as follows.

${\mathit{\Psi }}_{\mathbf{n}}\mathbf{=}\sqrt{\frac{\mathbf{2}}{\mathbf{L}}}\mathit{s}\mathit{i}\mathit{n}\left(\frac{n\pi }{L}x\right)$ …(i)

Step 3: calculate the effect of excitation on the probability in the small length of x axis at the centre of the well.

The probability of detecting the electron is given by,

$P_n=|{\Psi }_n{|}^2$

Substitute $\sqrt{\frac{2}{L}}\sin \left(\frac{n\pi }{L}x\right)$for ${\Psi }_n$into above equation.

    $P_n=\frac{2}{L}{sin}^2\left(\frac{n\mathrm{\pi }}{L}x\right)$                                                                                              …(ii)

Initially electron is in the ground state, therefore$n=1$

Substitute$1$ for $n$into equation (ii).

$P_1=\frac{2}{L}{sin}^2\left(\frac{\mathrm{\pi }}{L}X\right)$

When electron is excited from ground state to first state excited, then $n=2$.

Substitute2 for $n$into equation (ii).

$P_2=\frac{2}{L}{sin}^2\left(\frac{2\mathrm{\pi }}{L}X\right)$

 

To check the probability, divide the probability $P_1$by $P_2$.

 $$\begin{gathered} \frac{P_2}{P_1}=\frac{{\displaystyle \frac{2}{L}}{\sin }^2\left({\displaystyle \frac{2\pi }{L}}x\right)}{\frac{2}{L}{\sin }^2\left(\frac{2\pi }{L}x\right)} \\ \frac{P_2}{P_1}=\frac{{\sin }^2\left(\frac{2\pi }{L}x\right)}{{\sin }^2\left(\frac{2\pi }{L}x\right)} \end{gathered}$$                                                                                                …(iii)

Let assume point near the centre,$x=0.51L$ .

$$\begin{gathered} \frac{P_2}{P_1}=\frac{{\sin }^2\left(\frac{2\pi }{L}\times 0.51L\right)}{{\sin }^2\left(\frac{2\pi }{L}\times 0.51L\right)} \\ \frac{P_2}{P_1}=\frac{{\sin }^2\left(1.01\pi \right)}{{\sin }^2\left(1.51\pi \right)} \\ \frac{P_2}{P_1}<1 \end{gathered}$$

 

From the above discussion, the probability is decreases as the electron go from the ground state to first excited state at the centre.

Step 4: calculate the effect of excitation on the probability in the small length of x axis near one the wellwalls.

Let assume the point near the well walls,$x=0.99L$

Substitute$0.99L$ for $x$into equation (iii).

$$\begin{gathered} \frac{P_2}{P_1}=\frac{{\sin }^2\left(\frac{2\pi }{L}\times 0.99L\right)}{{\sin }^2\left(\frac{2\pi }{L}\times 0.99L\right)} \\ \frac{P_2}{P_1}=\frac{{\sin }^2\left(1.98\pi \right)}{{\sin }^2\left(0.99\pi \right)} \\ \frac{P_2}{P_1}>1 \end{gathered}$$

 

From the above discussion, the probability is increases as the electron go from the ground state to first excited near the one of the well walls.